Scattering Amplitudes

Transcription

1 Scattering Amplitudes LECTURE 1 Jaroslav Trnka Center for Quantum Mathematics and Physics (QMAP), UC Davis ICTP Summer School, June 2017

2 Particle experiments: our probe to fundamental laws of Nature

3 Theorist s perspective: scattering amplitude Initial states Final states

4 Phenomenology Initial states Tool how to learn about the dynamics: interactions, theories, symmetries Final states

5 What does the blob really represent?

6 What does the blob really represent? but there is more than that..

7 It can be for example a sum of different pictures _ _ _

8 And in a special case even something more surprising

9 Overview of lectures Lecture 1: Review of scattering amplitudes Motivation On-shell amplitudes Kinematics of massless particles Lecture 2: New methods for amplitudes Recursion relations for tree-level amplitudes Unitarity methods for loop amplitudes On-shell diagrams Lecture 3: Geometric formulation Toy model: N=4 SYM theory Positive Grassmannian Amplituhedron

10 Motivation

11 Quantum Field Theory (QFT) Our theoretical framework to describe Nature Compatible with two principles Special relativity Quantum mechanics

12 Perturbative QFT (Dirac, Heisenberg, Pauli; Feynman, Dyson, Schwinger) Fields, Lagrangian, Path integral L = 1 4 F µ F µ + i 6D m Z is(a,,,j) DA D D e Feynman diagrams: pictures of particle interactions Perturbative expansion: trees, loops

13 Great success of QFT QFT has passed countless tests in last 70 years Example: Magnetic dipole moment of electron 1928 Theory: Experiment: g e =2 g e 2

14 Great success of QFT QFT has passed countless tests in last 70 years Example: Magnetic dipole moment of electron 1947 Theory: Experiment: g e = g e =2.0023

15 Great success of QFT QFT has passed countless tests in last 70 years Example: Magnetic dipole moment of electron 1957 Theory: Experiment: g e = g e =

16 Great success of QFT QFT has passed countless tests in last 70 years Example: Magnetic dipole moment of electron 1990 Theory: Experiment: g e = g e =

17 Dualities At strong coupling: perturbative expansion breaks Surprises: dual to weakly coupled theory Gauge-gauge dualities (Montonen-Olive 1977, Seiberg-Witten 1994) Gauge-gravity duality (Maldacena 1997)

18 Incomplete picture Our picture of QFT is incomplete Also, tension with gravity and cosmology If there is a new way of thinking about QFT, it must be seen even at weak coupling Explicit evidence: scattering amplitudes

19 Colliders at high energies Proton scattering at high energies LHC - gluonic factory Needed: amplitudes of gluons for higher multiplicities gg! gg... g

20 Early 80s Status of the art: gg! ggg Brute force calculation 24 pages of result (k 1 k 4 )( 2 k 1 )( 1 3 )( 4 5 )

21 New collider 1983: Superconducting Super Collider approved Energy 40 TeV: many gluons! Demand for calculations, next on the list: gg! gggg

22 Parke-Taylor formula (Parke, Taylor 1985) Process gg! gggg 220 Feynman diagrams, 100 pages of calculations 1985: Paper with 14 pages of result

23 Parke-Taylor formula Process gg! gggg 220 Feynman diagrams, 100 pages of calculations

24 Parke-Taylor formula

25 Parke-Taylor formula Within a year they realized M 6 = h12i 3 h23ih34ih45ih56ih61i Spinor-helicity variables p µ = µ aȧ a ȧ (1) h12i = ab a [12] = ȧḃ (1) ȧ (2) (2) b ḃ (Mangano, Parke, Xu 1987)

26 Parke-Taylor formula Within a year they realized AN AMPLITUDE FOR n GLUON SCATTERING M n = h12i 3 h23ih34ih45i...hn1i STEPHEN 3. PARKE and T. R. TAYLOR Fermi National Accelerator Laboratory P.O. Box 500, Batavia, IL

27 Problems with Feynman diagrams Particles on internal lines are not real Individual diagrams not gauge invariant Obscure simplicity of the final answer Most of the terms in each diagram cancels Lesson: work with gauge invariant quantities with fixed spin structure

28 Birth of amplitudes New field in theoretical particle physics New methods and effective calculations Uncovering new structures in QFT Road map Explicit calculation New structure discovered New method which exploits it

29 What are scattering amplitudes

30 Scattering process Interaction of elementary particles Initial state ii and final state fi Scattering amplitude M if = hi fi Example: e + e! e + e or e + e! etc. Cross section: Z = M 2 d probability

31 Scattering amplitude in QFT Scattering amplitude depends on types of particles and their momenta M if = F (p i,s i ) Theoretical framework: calculated in some QFT Specified by Lagrangian: interactions and couplings L = L(O j,g k ) Example: QED L int = e µ A µ

32 Perturbation theory Weakly coupled theory M = M 0 + g M 1 + g 2 M 2 + g 3 M Representation in terms of Feynman diagrams Perturbative expansion = loop expansion M = M tree + M 1 loop + M 2 loop +...

33 Divergencies Loop diagrams are generally UV divergent Z 1 1 d 4` (`2 + m 2 )[(` + p) 2 + m 2 ] log IR divergencies: physical effects, cancel in cross section Dimensional regularization: calculate integrals in 4+ dimensions Divergencies 1 k

34 Renormalizable theories Absorb UV divergencies: counter terms Finite number of them: renormalizable theory Infinite number: non-renormalizable theory Mostly only renormalizable theories are interesting Exceptions: effective field theories Example: Chiral perturbation theory - derivative expansion L = L 2 + L 4 + L 6 + L Different loop orders are mixed

35 Analytic structure of amplitudes Tree-level: rational functions g 2 (p 1 + p 2 ) 2 Only poles Loops: polylogarithms and more complicated functions log 2 (s/t) Branch cuts

36 Kinematics of massless particles

37 Massless particles Parameters of elementary particles of spin S Spin s =( S, S) On-shell (physical) particle Mass m Momentum p µ p 2 = m 2 Massless particle: m =0 p 2 =0 spin = helicity: only two extreme values h = { S, S} Example: photon h =(+, ) s =0 missing

38 Spin functions At high energies particles are massless Fundamental laws reveal there Spin degrees of freedom: spin function s=0: Scalar - no degrees of freedom s=1/2: Fermion - spinor u s=1: Vector - polarization vector s=2: Tensor - polarization tensor µ h µ

39 Spin functions At high energies particles are massless Fundamental laws reveal there Spin degrees of freedom: spin function s=0: Scalar - no degrees of freedom s=1/2: Fermion - spinor u s=1: Vector - polarization vector s=2: Tensor - polarization tensor µ h µ

40 Polarization vectors Spin 1 particle is described by vector µ 2 degrees of freedom 4 degrees of freedom Null condition: =0 3 degrees of freedom left We further impose: p =0 Identification Feynman diagrams depend on gauge dependence µ µ + p µ

41 Spinor helicity variables Standard SO(3,1) notation for momentum p µ =(p 0,p 1,p 2,p 3 ) p j 2 R We use SL(2,C) representation p ab = µ ab p µ = p0 + ip 1 p 2 + p 3 p 2 p 3 p 0 ip 1 On-shell: p 2 =det(p ab )=0 Rank (p ab )=1 p 2 = p p p 2 2 p 2 3

42 Spinor helicity variables We can then write p ab = a apple b SL(2,C): dotted notation p a ḃ = a e ḃ e is complex conjugate of Little group transformation! t e! 1 t e leaves momentum unchanged 3 degrees of freedom p! p

43 Spinor helicity variables Momentum invariant (p 1 + p 2 ) 2 =(p 1 p 2 ) p µ 1 = µ aȧ 1a e 1ȧ p µ 2 = µ bḃ 2b e 2ḃ Plugging for momenta (p 1 p 2 )=( µ aȧ µbḃ)( 1a 2b)( e 1ȧ e 2ḃ) ab ȧḃ =( ab 1a 2b )( ȧḃ e 1ȧ e 2ḃ) Define: h12i ab 1a 2b [12] ȧḃ e 1ȧ e 2ḃ

44 Invariant products Momentum invariant s ij =(p i + p j ) 2 = hiji[ij] Square brackets Angle brackets hiji = ab ia jb [ij] = ȧḃ e iȧ e jḃ Antisymmetry More momenta h21i = h12i [21] = [12] (p 1 + p 2 + p 3 ) 2 = h12i[12] + h23i[23] + h13i[13]

45 Invariant products Shouten identity Mixed brackets h13ih24i = h12ih34i + h14ih23i h ] h12i[24] + h13i[34] Momentum conservation nx i=1 ia e iȧ =0 Non-trivial conditions: Quadratic relation between components

46 Polarization vectors Two polarization vectors µ + = µ aȧ a eȧ h i µ = µ aȧ ae ȧ [e e ] Note that where, e are auxiliary spinors ( + )=1 Freedom in choice of, e µ µ + p µ corresponds to Gauge redundancy of Feynman diagrams

47 Scaling of amplitudes Consider some amplitude A( + + ) Little group scaling A =( n ) Q depends only on momenta! t e! 1 t e p! p +! 1 t 2 +! t 2 A(i )! t 2 A(i ) A(i + )! 1 t 2 A(i+ )

48 Back to Parke-Taylor formula Let us consider Scaling A( ) A t i, 1 e i = t 2 A( i, e i) t A t i, 1 e i = 1 t t 2 A( i, e i) for particles 1,2 for particles 3,4,5,6 Check for explicit expression h12i 3 A 6 = h23ih34ih45ih56ih61i If only hiji allowed the form is unique

49 Helicity amplitudes In Yang-Mills theory we have + or - gluons A 6 ( ) We denote k: number of - helicity gluons Some amplitudes are zero A n ( ) = 0 A n ( ++ +) = 0 A n ( )=0 A n (+ )=0 First non-trivial: k=2 A n ( + +) Parke-Taylor formula for tree level

50 Three point amplitudes

51 Three point kinematics Gauge invariant building blocks: on-shell amplitudes p 2 1 = p 2 2 = p 2 3 =0 p 1 + p 2 + p 3 =0 Plugging second equation into the first (p 1 + p 2 ) 2 =(p 1 p 2 )=0 Similarly we get for other pairs (p 1 p 2 )=(p 1 p 3 )=(p 2 p 3 )=0 These momenta are very constrained!

52 Three point kinematics Use spinor helicity variables trivializes on-shell condition p 1 = 1 e 1, p 2 = 2 e 2, p 3 = 3 e 3 The mutual conditions then translate to (p 1 p 2 )=h12i[12] = 0 And similarly for other two pairs (p 1 p 3 )=h13i[13] = 0 (p 2 p 3 )=h23i[23] = 0

53 Two solutions We want to solve conditions h12i[12] = h13i[13] = h23i[23] = 0 Solution 1: h12i =0which implies 2 = 1 Then we also have h23i = h13i And we set h13i =0 by demanding 3 =

54 Two solutions Solution 2: [12] = [23] = [13] = 0 e 1 e 2 e 3 Let us take this solution p 1 = 1 e 1, p 2 = 2 e 1, p 3 =( 1 2 ) e 1 complex momenta No solution for real momenta

55 Three point amplitudes Gauge theory: scattering of three gluons (not real) Building blocks: h12i, h23i, h13i, [12], [23], [13] Mass dimension: each term Three point amplitude m A 3 3 p p m

56 Three point amplitudes Two options A (1) 3 = h12i a 1 h13i a 2 h23i a 3 A (2) 3 = [12] b 1 [13] b 2 [23] b 3 Apply to A 3 (1, 2, 3 + ) A 3 (t 1,t 1 e 1 )=t a 1+a2 A 3 A 3 (t 2,t 1 e 2 )=t a 1+a3 A 3 A 3 (t 3,t 1 e 3 )=t a 2+a3 A 3 a 1 + a 2 =2 a 1 + a 3 =2 a 2 + a 3 = 2 a 1 =3 a 2 = 1 a 3 = 1

57 Three point amplitudes Similarly for A 3 (1 +, 2 +, 3 ) Two fundamental amplitudes A 3 (1, 2, 3 + )= h12i3 h13ih23i A 3 (1 +, 2 +, 3 )= [12]3 [13][23] This is true to all orders: just kinematics

58 Three point amplitudes Collect all ( +) amplitudes A 3 (1, 2, 3 + )= A 3 (1, 2 +, 3 )= A 3 (1 +, 2, 3 )= Similarly for h12i3 h13ih23i h13i3 h12ih23i h23i3 h12ih13i (+ + ) ) amplitudes habi 4 h12ih23ih31i where a,b are - helicity gluons [ab] 4 [12][23][31] where a,b are + helicity gluons

59 Three point amplitudes Using similar analysis we find for gravity habi 8 h12i 2 h23i 2 h31i 2 where a,b are - helicity gravitons [ab] 8 [12] 2 [23] 2 [31] 2 where a,b are + helicity gravitons Note that there is no three-point scattering They exist only for complex momenta Important input into on-shell methods

60 General 3pt amplitudes Two solutions for 3pt kinematics h 1 h 2 h 3 h 1 h 2 h Under the little group rescaling: A 3 (t j,t 1 e j ) t 2hj A 3 e 1 e 2 e 3 Solve the system of equations

61 General 3pt amplitudes Two solutions for amplitudes h 1 h 2 A 3 = [12] h 1 h 2 +h 3 [23] h 2 h 3 +h 1 [31] h 1 h 3 +h 2 h 3 h 1 h 2 h 3 h A 3 = h12i 1 +h 2 h 3 h23i h 2+h 3 h 1 h31i h 1+h 3 h 2 Which one is correct?

62 General 3pt amplitudes Two solutions for amplitudes h 1 h 2 A 3 = [12] +h 1+h 2 h 3 [23] h 1 +h 2 +h 3 [31] +h 1 h 2 +h 3 h 3 h 1 + h 2 + h 3 apple 0 h 1 h 2 h 3 A 3 = h12i h 1 h 2 +h 3 h23i +h 1 h 2 h 3 h31i h 1+h 2 h 3 h 1 + h 2 + h 3 0 Mass dimension must be positive!

63 All spins allowed Note that these formulas are valid for any spins For example for amplitude A 3 (1 0, 2 1+, 3 2+ ) A 3 = h23ih31i3 h12i 3 But we can also do higher spins A 3 = h23i10 h31i 14 h12i 20 Completely fixed just by kinematics! A 3 (1 3+, 2 5+, 3 12 )

64 Tree-level amplitudes

65 Feynman diagrams Yang-Mills Lagrangian L = 1 4 F µ F µ (@A) 2 + A + A 4 f abc g µ p f abe f cde g µ g Draw diagrams Feynman rules Sum everything

66 Change of strategy What is the scattering amplitude? Feynman diagrams Unique object fixed by physical properties Was not successful (1960s) Modern methods use both: Calculate the amplitude directly Use perturbation theory

67 Locality and unitarity Only poles: Feynman propagators Locality 1 P 2 X where P = k2p p k On the pole Unitarity M! P 2 =0 M L 1 P 2 M R Feynman diagrams recombine on both sides into amplitudes

68 Factorization on the pole On P 2 =0 Res M = M L 1 P 2 M R For P 2 =0 the internal leg: on-shell physical particle Both sub-amplitudes are on-shell, gauge invariant On-shell data: statement about on-shell quantities

69 On-shell constructibility Factorization of tree-level amplitudes M! P 2 =0 M L 1 P 2 M R On-shell constructibility: factorizations fix the answer Write a proposal tree-level amplitude On-shell gauge invariant function, correct weights It factorizes properly on all channels The amplitude is uniquely specified by these properties fm

70 On-shell constructibility This is obviously a theory specific statement Theories with contact terms might not be constructible Naively, this is false for Yang-Mills theory gg! gg Four point amplitude Contact term

71 On-shell constructibility This is obviously a theory specific statement Theories with contact terms might not be constructible Naively, this is false for Yang-Mills theory gg! gg Four point amplitude Contact term Imposing gauge invariance fixes it

72 On-shell constructibility In gravity we have infinity tower of terms Only h 3 L p gr h 2 + h 3 + h terms important, others fixed by diffeomorphism symmetry On-shell constructibility of Yang-Mills, GR, SM Only function which factorizes properly on all poles is the amplitude.

73 Four point test

74 From 3pt to 4pt Three point amplitudes exist for all spins For 4pt amplitude: we have a powerful constraint A 4! s=0 A 3 1 s A 3 This must be true on all channels This will immediately kill most of the possibilities We are left with spectrum of spins: 0, 1 2, 1, 3 2, 2

75 Three point of spin S I will discuss amplitudes of single spin S particle For 3pt amplitudes we get h12i 3 S [12] 3 A 3 = A 3 = h23ih31i minimal [23][31] powercounting ( +) (+ + ) S There exist also non-minimal amplitudes A 3 =(h12ih23ih31i) S ( ) A 3 = ([12][23][31]) S (+ + +)

76 Four point amplitude Let us consider a 4pt amplitude of particular helicities Mandelstam variables: A 4 ( ++) One can show that the little group dictates: A 4 =(h12i[34]) 2S F (s, t) It must be consistent with factorizations s =(p 1 + p 2 ) 2 = h12i[12] = h34i[34] t =(p 1 + p 4 ) 2 = h14i[14] = h23i[23] u =(p 1 + p 3 ) 2 = h13i[13] = h24i[24]

77 s-channel constraint The s-channel factorization dictates P =1+2= S P + 3 +S P 1 S 4 +S h12i 3 S 1 [34] 3 S A 4! on s=0 h1p ih2p i s [3P ][4P ] Note: s = h12i[12] = h34i[34]

78 s-channel constraint The s-channel factorization dictates P =1+2= S P + 3 +S P 1 S 4 +S h12i 3 S 1 [34] 3 S A 4! on s=0 h1p ih2p i s [3P ][4P ] Note: s = h12i[12] = h34i[34]

79 s-channel constraint h12i 3 S 1 [34] 3 S h1p ih2p i s [3P ][4P ] Rewrite using momentum conservation: h1p i[3p ]= h1 P 3] = h ] = h12i[23] h2p i[4p ]= h2 P 4] = h ] = h23i[34] We get 1 (h12i[34]) 3 S s h1p i[3p ]h2p i[4p ]

80 s-channel constraint h12i 3 S 1 [34] 3 S h1p ih2p i s [3P ][4P ] Rewrite using momentum conservation: h1p i[3p ]= h1 P 3] = h ] = h12i[23] h2p i[4p ]= h2 P 4] = h ] = h23i[34] We get 1 (h12i[34]) 3 S s h12i[23]h23i[34]

81 s-channel constraint h12i 3 S 1 [34] 3 S h1p ih2p i s [3P ][4P ] Rewrite using momentum conservation: h1p i[3p ]= h1 P 3] = h ] = h12i[23] h2p i[4p ]= h2 P 4] = h ] = h23i[34] We get 1 (h12i[34]) 2 S s h23i[23]

82 s-channel constraint h12i 3 S 1 [34] 3 S h1p ih2p i s [3P ][4P ] Rewrite using momentum conservation: h1p i[3p ]= h1 P 3] = h ] = h12i[23] h2p i[4p ]= h2 P 4] = h ] = h23i[34] We get 1 (h12i[34]) 2 S s t

83 s-channel constraint h12i 3 S 1 [34] 3 S h1p ih2p i s [3P ][4P ] Rewrite using momentum conservation: h1p i[3p ]= h1 P 3] = h ] = h12i[23] h2p i[4p ]= h2 P 4] = h ] = h23i[34] We get (h12i[34]) 2S 1 st S Note: t = u Trivial helicity factor Important piece

84 Comparing channels On s-channel we got: On t-channel we would get: A 4! (h12i[34]) 2S A 4! (h12i[34]) 2S 1 st S 1 ts S Require simple poles: 1 and search for F (s, t, u) s, 1 t, 1 u

85 Comparing channels On s-channel we got: A 4! (h12i[34]) 2S 1 st S On t-channel we would get: A 4! (h12i[34]) 2S 1 ts S Require simple poles: There are only two solutions: F (s, t, u) = 1 s + 1 t + 1 u 3 1 s, 1 t, 1 u and search for F (s, t, u) = 1 stu spin 0 ( ) spin 2 (GR) F (s, t, u)

86 Where are gluons (spin-1)? Need to consider multiplet of particles h12i 3 S A 3 = [12] 3 S f a 1a 2 a 3 A 3 = h23ih31i f a 1a 2 a 3 [23][31] The same check gives us S=1 and requires f a 1a 2 a P f a 3a 4 a P + f a 1a 4 a P f a 2a 3 a P = f a 1a 3 a P f a 2a 4 a P and the result corresponds to SU(N) Yang-Mills theory

87 Power of 4pt check We can apply this check for cases with mixed particle content: Spin >2 still not allowed Spin 2 is special: only one particle and it couples universally to all other particles We get various other constraints on interactions (of course all consistent with known theories) General principles very powerful

88 Thank you for attention!